Type I errors Problem set A One question at a time

Find the probability of making a Type I error in each test.

1. (a) X ~ N(μ, 5.2²), n = 14, 2% significance level; H0: μ = 120, H1: μ > 120.

   • The distribution of the sample mean is continuous, so P(Type I error) is the significance level.
   • P(Type I error) = 0.02

2. (b) X ~ N(μ, 0.8²), n = 25, 8% significance level; H0: μ = 3.5, H1: μ ≠ 3.5.

   • The distribution of the sample mean is continuous, so P(Type I error) is the significance level.
   • P(Type I error) = 0.08

X is normally distributed with standard deviation σ. A sample of size n is used to test H0: μ = μ0. Find the probability of making a Type I error for each critical region.

1. (a) σ = 4.5, n = 9, H0: μ = 52, H1: μ < 52; critical region \(\bar X < 50\).

   • Under H0, X̄ ~ N(52, 4.5²/9), standard deviation 1.50
   • P(Type I error) = P(X̄ < 50) = 0.0912

2. (b) σ = 2, n = 16, H0: μ = 10, H1: μ ≠ 10; critical region \(\bar X < 9.2\) or \(\bar X > 10.8\).

   • Under H0, X̄ ~ N(10, 2²/16), standard deviation 0.500
   • P(Type I error) = P(X̄ < 9.2) + P(X̄ > 10.8)
   • = 0.0548 + 0.0548 = 0.110

X ~ B(n, p). Find the probability of making a Type I error for each test.

1. (a) n = 15, H0: p = 0.6, H1: p < 0.6; critical region X ≤ 5.

   • Under H0, X ~ B(15, 0.6)
   • From the calculator: P(Type I error) = P(X ≤ 5) = 0.0338

2. (b) n = 10, H0: p = 0.3, H1: p > 0.3; critical region X ≥ 6.

   • Under H0, X ~ B(10, 0.3)
   • P(Type I error) = P(X ≥ 6) = 1 − P(X ≤ 5)
   • From the calculator: 1 − 0.95265 = 0.0473

X ~ Po(λ). Find the probability of making a Type I error for each test.

1. (a) H0: λ = 5.5, H1: λ < 5.5; critical region X ≤ 2.

   • Under H0, X ~ Po(5.5)
   • From the calculator: P(Type I error) = P(X ≤ 2) = 0.0884

2. (b) H0: λ = 4.1, H1: λ > 4.1; critical region X ≥ 9.

   • Under H0, X ~ Po(4.1)
   • P(Type I error) = P(X ≥ 9) = 1 − P(X ≤ 8)
   • From the calculator: 1 − 0.97551 = 0.0245

A factory says that 8% of its bolts are faulty. A customer thinks that the true proportion is higher. He tests a random sample of 40 bolts at the 5% significance level.

1. (a) Find the critical region for the test.

   • H0: p = 0.08, H1: p > 0.08. Under H0, X ~ B(40, 0.08).
   • From the calculator: P(X ≥ 6) = 0.0967 > 0.05 and P(X ≥ 7) = 0.0376 ≤ 0.05
   • Critical region: X ≥ 7

2. (b) Find the probability of making a Type I error.

   • P(Type I error) = P(X ≥ 7) = 0.0376
   • This is less than 5%, because X takes whole values only.

The number of customers who arrive at a kiosk in 10 minutes is modelled by a Poisson distribution with mean 11. The owner thinks that a new sign has increased this mean. She will count the arrivals in one 10-minute period and test, at the 5% significance level.

1. (a) Find the critical region for the test.

   • H0: λ = 11, H1: λ > 11. Under H0, X ~ Po(11).
   • From the calculator: P(X ≥ 17) = 0.0559 > 0.05 and P(X ≥ 18) = 0.0322 ≤ 0.05
   • Critical region: X ≥ 18

2. (b) Find the probability of making a Type I error.

   • P(Type I error) = P(X ≥ 18) = 1 − P(X ≤ 17) = 1 − 0.96781 = 0.0322
   • This is less than 5%.

A bakery makes rolls with a mean mass of 80 g and a standard deviation of 6 g. An inspector takes a random sample of 16 rolls and rejects the batch if the sample mean is below 77.5 g. Assume that masses are normally distributed. Find the probability that a batch with mean mass 80 g is wrongly rejected.

• This is a Type I error: H0 (mean 80) is true but is rejected.
• Under H0, X̄ ~ N(80, 6²/16), standard deviation 1.50
• P(X̄ < 77.5) = 0.0478

A dairy states that the mean fat content of its milk is 3.5%. An inspector tests, at the 5% significance level, whether the mean fat content is different from 3.5%. She uses the mean of a random sample of 20 cartons.

1. (a) Describe a Type I error in this context.

   • A Type I error is rejecting H0 when H0 is true.
   • H0: the mean fat content is 3.5%. Rejecting it when it is true means wrongly deciding that the mean is different.

2. (b) Find the probability of making a Type I error.

   • The sample mean has a continuous distribution, so P(Type I error) is the significance level.
   • P(Type I error) = 0.05

A drug is said to work for 30% of patients. A doctor thinks that it works for fewer, and tests this at the 5% significance level with a random sample of 25 patients.

1. (a) Find the critical region for the test.

   • H0: p = 0.3, H1: p < 0.3. Under H0, X ~ B(25, 0.3).
   • From the calculator: P(X ≤ 3) = 0.0332 ≤ 0.05 and P(X ≤ 4) = 0.0905 > 0.05
   • Critical region: X ≤ 3

2. (b) Find the probability of making a Type I error.

   • P(Type I error) = P(X ≤ 3) = 0.0332

3. (c) Explain why this probability is not exactly 0.05.

   • The next value would give 0.0905, which is above 0.05.
   • So the actual probability 0.0332 is below the nominal level of 5%.

A teacher says that the mean score on a test is 70, with standard deviation 12. A researcher decides to reject this claim in favour of a higher mean if the sample mean is above 73. Assume that scores are normally distributed.

1. (a) Find the probability of a Type I error when the sample size is 25.

   • Under H0, X̄ ~ N(70, 12²/25), standard deviation 2.40
   • P(Type I error) = P(X̄ > 73) = 0.106

2. (b) Find the probability of a Type I error when the sample size is 100.

   • Under H0, X̄ ~ N(70, 12²/100), standard deviation 1.20
   • P(Type I error) = P(X̄ > 73) = 0.00621

3. (c) Explain the change.

   • A larger sample makes the sample mean less spread out around the H0 mean.
   • The same critical value is further away in standard deviations, so the probability falls from 0.106 to 0.00621.